Lever Rule

Materials Science · FE Reference Handbook section

Materials Science

5 formulas

10 exam-style examples

~55 min

All Materials Science lectures

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The following phase diagram and equations illustrate how the weight of each phase in a two-phase system can be determined:

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Lever rule phase fractions — Lever Rule

An alloy of overall composition 42 wt% B lies in a two-phase field bounded by 13 wt% and 83 wt% B. What fraction of each phase is present?

Given

$C_{0} = 42 wt%$
$C_L = 13 wt%$
$C_S = 83 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (83 - 42)/(83 - 13) = 0.586$
Solid fraction
$W_S = 1 - 0.586 = 0.414$
Check
$58.6% + 41.4% = 100% ✓$

Answer:

≈ 58.6% liquid, 41.4% solid

Why the other options are there

41.4% liquid (arms swapped)
42% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Lever Rule

Example 2

Lever rule phase fractions — Lever Rule (2)

An alloy of overall composition 50 wt% B lies in a two-phase field bounded by 25 wt% and 81 wt% B. What fraction of each phase is present?

Given

$C_{0} = 50 wt%$
$C_L = 25 wt%$
$C_S = 81 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (81 - 50)/(81 - 25) = 0.554$
Solid fraction
$W_S = 1 - 0.554 = 0.446$
Check
$55.4% + 44.6% = 100% ✓$

Answer:

≈ 55.4% liquid, 44.6% solid

Why the other options are there

44.6% liquid (arms swapped)
50% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Lever Rule

Example 3

Lever rule phase fractions — Lever Rule (3)

An alloy of overall composition 42 wt% B lies in a two-phase field bounded by 22 wt% and 82 wt% B. What fraction of each phase is present?

Given

$C_{0} = 42 wt%$
$C_L = 22 wt%$
$C_S = 82 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (82 - 42)/(82 - 22) = 0.667$
Solid fraction
$W_S = 1 - 0.667 = 0.333$
Check
$66.7% + 33.3% = 100% ✓$

Answer:

≈ 66.7% liquid, 33.3% solid

Why the other options are there

33.3% liquid (arms swapped)
42% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Lever Rule

Example 4

Lever rule phase fractions — Lever Rule (4)

An alloy of overall composition 51 wt% B lies in a two-phase field bounded by 13 wt% and 70 wt% B. What fraction of each phase is present?

Given

$C_{0} = 51 wt%$
$C_L = 13 wt%$
$C_S = 70 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (70 - 51)/(70 - 13) = 0.333$
Solid fraction
$W_S = 1 - 0.333 = 0.667$
Check
$33.3% + 66.7% = 100% ✓$

Answer:

≈ 33.3% liquid, 66.7% solid

Why the other options are there

66.7% liquid (arms swapped)
51% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Lever Rule

Example 5

Lever rule phase fractions — Lever Rule (5)

An alloy of overall composition 47 wt% B lies in a two-phase field bounded by 15 wt% and 71 wt% B. What fraction of each phase is present?

Given

$C_{0} = 47 wt%$
$C_L = 15 wt%$
$C_S = 71 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (71 - 47)/(71 - 15) = 0.429$
Solid fraction
$W_S = 1 - 0.429 = 0.571$
Check
$42.9% + 57.1% = 100% ✓$

Answer:

≈ 42.9% liquid, 57.1% solid

Why the other options are there

57.1% liquid (arms swapped)
47% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Lever Rule

Example 6

Lever rule phase fractions — Lever Rule (6)

An alloy of overall composition 40 wt% B lies in a two-phase field bounded by 14 wt% and 76 wt% B. What fraction of each phase is present?

Given

$C_{0} = 40 wt%$
$C_L = 14 wt%$
$C_S = 76 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (76 - 40)/(76 - 14) = 0.581$
Solid fraction
$W_S = 1 - 0.581 = 0.419$
Check
$58.1% + 41.9% = 100% ✓$

Answer:

≈ 58.1% liquid, 41.9% solid

Why the other options are there

41.9% liquid (arms swapped)
40% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Lever Rule

Example 7

Lever rule phase fractions — Lever Rule (7)

An alloy of overall composition 48 wt% B lies in a two-phase field bounded by 17 wt% and 78 wt% B. What fraction of each phase is present?

Given

$C_{0} = 48 wt%$
$C_L = 17 wt%$
$C_S = 78 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (78 - 48)/(78 - 17) = 0.492$
Solid fraction
$W_S = 1 - 0.492 = 0.508$
Check
$49.2% + 50.8% = 100% ✓$

Answer:

≈ 49.2% liquid, 50.8% solid

Why the other options are there

50.8% liquid (arms swapped)
48% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Lever Rule

Example 8

Lever rule phase fractions — Lever Rule (8)

An alloy of overall composition 36 wt% B lies in a two-phase field bounded by 18 wt% and 78 wt% B. What fraction of each phase is present?

Given

$C_{0} = 36 wt%$
$C_L = 18 wt%$
$C_S = 78 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (78 - 36)/(78 - 18) = 0.700$
Solid fraction
$W_S = 1 - 0.700 = 0.300$
Check
$70.0% + 30.0% = 100% ✓$

Answer:

≈ 70.0% liquid, 30.0% solid

Why the other options are there

30.0% liquid (arms swapped)
36% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Lever Rule

Example 9

Lever rule phase fractions — Lever Rule (9)

An alloy of overall composition 50 wt% B lies in a two-phase field bounded by 12 wt% and 85 wt% B. What fraction of each phase is present?

Given

$C_{0} = 50 wt%$
$C_L = 12 wt%$
$C_S = 85 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (85 - 50)/(85 - 12) = 0.479$
Solid fraction
$W_S = 1 - 0.479 = 0.521$
Check
$47.9% + 52.1% = 100% ✓$

Answer:

≈ 47.9% liquid, 52.1% solid

Why the other options are there

52.1% liquid (arms swapped)
50% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Lever Rule

Example 10

Lever rule phase fractions — Lever Rule (10)

An alloy of overall composition 32 wt% B lies in a two-phase field bounded by 16 wt% and 73 wt% B. What fraction of each phase is present?

Given

$C_{0} = 32 wt%$
$C_L = 16 wt%$
$C_S = 73 wt%$

Find

Phase fractions

Start with the thinking

The lever rule uses the opposite arm of the tie line.
Fractions must sum to 1.

Step-by-step solution

Liquid fraction
$W_L = (C_S - C_{0})/(C_S - C_L)$
Substituting
$W_L = (73 - 32)/(73 - 16) = 0.719$
Solid fraction
$W_S = 1 - 0.719 = 0.281$
Check
$71.9% + 28.1% = 100% ✓$

Answer:

≈ 71.9% liquid, 28.1% solid

Why the other options are there

28.1% liquid (arms swapped)
32% liquid (composition read as a fraction)

Reference: FE Reference Handbook — Materials Science → Lever Rule

Lever Rule

Handbook notes for this section

Core formulas for this FE topic

β+L\beta+Lβ+L

α+L\alpha+Lα+L

(Indiagram,L=liquid.)Ifx=theaveragecompositionattemperatureT,then(In diagram, L = liquid.) If x = the average composition at temperature T, then(Indiagram,L=liquid.)Ifx=theaveragecompositionattemperatureT,then

wt % a = x b x # 100Needs technical review

wt % b = x x # 100Needs technical review

Worked exam-style examples